1
Experiment
Did the subjects make choices “as if” they had a
preference relation  over bundles of (IC, HB)? If so,
could we infer  and predict future choices or offer
advice about choices?
In situation 2 the amount of money was $3.00 and the
prices were pHB = .50 and pIC = 1.00; in situation 5
the amount of money was $3.60 and the prices were
pHB = .60 and pIC = 1.20. The affordable set was
the same in these two cases. So if the framing of the
question doesn’t matter would expect the same choice
in 2 as in 5.
22% of the subjects did not make the same choice in
these two situations.
So we observe x  y and y  x for these people.
2
Lets look at situation 4 versus situation 1. In situation
4 the amount of money was $4.20 and the prices were
pHB = .80 and pIC = 1.20; in situation 1 the amount
of money was $3.60 and the prices were pHB = .40 and
pIC = 1.60. The affordable sets in these two cases are
graphed below.
If we observe choices x at 4 and y at 1 then we have
x  y and y  x. No one made choices like this.
3
Static Decision Theory Under Certainty
A set of objects X.
An individual is asked to express his preferences among
these objects or is asked to make choices from subsets
of X.
For x, y ∈ X we can ask which, if either, is strictly
preferred.
• If the individual says x is strictly better than y we
write x  y, read as x is strictly preferred to y.
• Â is a binary relation on X.
Example 1: X = {a, b, p}, b  a, a  p and b  p.
What if the answers also included a  b?
4
Axioms
Asymmetry: For any x, y ∈ X if x  y then
not[y  x].
Negative Transitivity: For any x, y, z ∈ X if
not[x  y] and not[y  z] then not[x  z].
Proposition. The binary relation  is negatively
transitive iff x  z implies that, for all y ∈ X, x  y or
y  z.
Example 2: X = {a, b, c}, b  a, a  c and b ? c. If we
have asymmetry and NT you also know how b and c
must be ranked.
Definition. A binary relation  is called a (strict)
preference relation if it is asymmetric and negatively
transitive.
Is Asymmetry a good normative or descriptive property? What about NT?
5
Weak Preference
Definition. For x, y ∈ X:
1. x is weakly preferred to y, x º y, if not[y  x].
2. x is indifferent to y, x ∼ y, if not[x  y] and
not[y  x].
Does the absence of strict preference in either
direction require real indifference or could it permit
non-comparability?
Example. X = {a, b, c}. Suppose a is not ranked (by
Â) relative to either b or c. If  satisfies NT, then b
and c are not ranked either.
An interesting alternative would be to ask about Â
and ∼ separately. Then define x º y as either x  y
or x ∼ y. This permits the possibility that x and y are
not comparable.
6
Definition. The binary relation º on X is complete if
for all x, y ∈ X, x º y, y º x or both. It is transitive if
x º y and y º z implies x º z.
Proposition. Let  be a binary relation on X.
1. Â is asymmetric iff º is complete.
2. Â is negatively transitive iff º is transitive.
Proof of ⇒
1. By asymmetry of  there is no pair x, y ∈ X such
that both x  y and y  x. So at least one of
not[x  y] and not[y  x] is true. Thus for any
x, y ∈ X either y º x or x º y or both. This is
completeness.
2. Using the definition of º, negative transitivity of
 is: for any x, y, z ∈ X, y º x and z º y implies
z º x. This is transitivity.
⇐ will be on homework 1.
7
Transitivity
Why do we care about transitivity?
Remark: If  is a preference relation then  is
transitive.
Normative property?
Important for choice.
Example. X = {a, b, p}. Consider a sequence of
choices from among pairs.
1. {a, b}, a  b and a is chosen.
2. {a, p}, p  a and p is chosen.
3. {p, b}, b  p and b is chosen.
4. {a, b} . . .
Without transitivity can get cycles.
Remark: If  is a preference relation then  is acyclic,
i.e. [x1 Â x2 Â . . . xn−1 Â xn ] ⇒ [x1 6= xn ].
8
Choice
Extend binary comparisons to choice over a set of more
objects.
A finite set of objects X. Let P (X) be the set of all
non-empty subsets of X.
Definition. For  a preference relation on X define
c(·, Â) by, for A ∈ P (X),
c(A, Â) = {x ∈ A : for all y ∈ A, y 6Â x}.
Interpretation: c(A, Â) is the set of alternatives chosen
from A by a decision maker with preferences Â.
Remark: If x, y ∈ c(A, Â) then x ∼ y.
Proposition. For  a preference relation on a finite
set X,
c(·, Â) : P (X) → P (X).
9
What else do we know about c(·, A)?
Consider general choice functions and ask what is
special about c(·, A).
Definition. A choice function for X is a function
c : P (X) → P (X) such that for all A ∈ P (X),
c(A) ⊂ A.
Clearly, c(·, Â) is a choice function.
Can any choice function be generated by some preference relation Â? No.
Example. X = {a, b, c}.
1. c({a, b, c}) = {a} and c({a, b}) = {b} ⇒ a violation
of asymmetry.
2. c({a, b}) = {a, b} and c({a, b, c}) = {b} ⇒ a
violation of NT.
10
Axioms
A
B
.x
Sen’s α. If x ∈ B ⊂ A and x ∈ C(A), then x ∈ C(B).
Independence of Irrelevant Alternatives.
Proposition. If  is a preference relation then c(·, Â)
satisfies Sen’s α.
Proof. Suppose there are sets A, B ∈ P (X) with
B ⊂ A, x ∈ c(A, Â) and x 6∈ c(B, Â). Then there is a
y ∈ B such that y  x. Since B ⊂ A we have y ∈ A
and y  x. Thus x 6∈ c(A, Â). A contradiction.
11
B
A
.
y
.x
Sen’s β. If x, y ∈ c(A), A ⊂ B and y ∈ c(B) then x ∈
C(B).
Proposition. If  is a preference relation then c(·, Â)
satisfies Sen’s β.
Proof. Since x ∈ c(A, Â) and y ∈ A we have y 6Â x. By
definition, y ∈ c(B, Â) implies that for all z ∈ B, z 6Â y.
By negative transitivity, y 6Â x and z 6Â y implies z 6Â x.
Since x ∈ B and this holds for all z ∈ B we have
x ∈ c(B, Â).
12
Are there any other restrictions on c(·, Â) that follow
from  being a preference relation? No.
Proposition. If a choice function c satisfies Sen’s α
and β, then there is a preference relation  such that
c(·) = c(·, Â).
Define the “revealed preference” relation  by
x  y if x 6= y and c({x, y}) = {x}.
To prove the proposition we need to show that  is a
preference relation and that c(·) = c(·, Â).
13
Proof
To show that  is a preference relation we need to show
that it is asymmetric and negatively transitive.
1. Asymmetry. Suppose for some x and y, that
x  y and y  x. Then c({x, y}) = {x} and
c({x, y}) = {y}. A contradiction.
2. Negative Transitivity. Suppose that for some
x, y, z ∈ X we have z 6Â y and y 6Â x. We need to
show that z 6Â x. This is x ∈ c({x, z}). By Sen’s α,
showing that x ∈ c({x, y, z}) is sufficient. Suppose
x 6∈ c({x, y, z}). Then at least one of y and z are
in c({x, y, z}).
Suppose y ∈ c({x, y, z}). Then by Sen’s α,
y ∈ c({x, y}). By y 6Â x we have x ∈ c({x, y}). By
Sen’s β this implies that x ∈ c({x, y, z}).
Suppose that z ∈ c({x, y, z}). Then by Sen’s α,
z ∈ c({y, z}). By z 6Â y we have y ∈ c({y, z}). By
Sens’ β this implies that y ∈ c({x, y, z}). By the
previous argument this implies that x ∈ c({x, y, z}).
14
We also need to show that for each A ∈ P (X),
c(A) = c(A, Â).
1. Suppose x ∈ c(A). Then by Sen’s α, x ∈ c({x, y})
for all y ∈ A. Thus for all y ∈ A, y 6Â x. So
x ∈ c(A, Â).
2. Suppose x ∈ c(A, Â). Then for all y ∈ A, y 6Â x.
So for all y ∈ A, x ∈ c({x, y}). Suppose x 6∈ c(A).
Then there is some z ∈ A, z 6= x such that z ∈ c(A).
By Sen’s α, z ∈ c({x, z}). Then c({x, z}) = {x, z},
{x, z} ⊂ A and z ∈ c(A). So by Sen’s β, x ∈ c(A).
A contradiction.
So we know,
[Sen’s α and β for c(·)] ⇔
[c(·) = c(·, Â) for the preference relation Â]
15
WARP
There is an alternative equivalent way to state Sen’s α
and β.
This is Houthaker’s Axiom which is also called the
Weak Axiom of Revealed Preference (WARP).
WARP: If x and y are both in A and B and if x ∈ c(A)
and y ∈ c(B), then x ∈ c(B) and y ∈ C(A).
Proposition. c(·) satisfies Sen’s α and β if and only if
it satisfies WARP.
16
Partial Orders
Completeness of º is questionable from both a descriptive and a normative point of view.
Definition. Â is a partial order if it is an asymmetric
and transitive binary relation.
We can define a choice function as before. What
properties does it have?
Sen’s α still holds, but Sen’s β may fail. (On homework
1.)
Now we would not want to define ∼ as before. x 6Â y and
y 6Â x could express indifference or non-comparability.
An alternative approach is to include a positive expression of indifference, i.e. preferences described by the
pair (Â, ∼).